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omeli [17]
3 years ago
8

Which of the following is NOT an allotrope of carbon? A. graphite B. carbon-13 C. a diamond D. a fullerene

Chemistry
1 answer:
egoroff_w [7]3 years ago
5 0
B. carbon-13 is not an allotrope of Carbon.
Allotropes<span> are elements on the periodic table that have more than one crystalline form. </span>Isotopes<span> are atoms of the same element with the same atomic number but have a different mass number.
C-13 is an isotope of carbon, not an allotrope.</span>
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Identify the electrophilic site in the following molecule, CH_3CH_2NHCH_2CH_3. (A) H (B) N (C) CH_2 (D) CH_3 (E) there is no ele
Serhud [2]

Answer:

(C) CH_2

Explanation:

An electrophile is an atom that loves electron. For an atom to be able to pull electrons closer to itself, it must be electron deficient and with a partial positive charge.

From the structure give, N is a highly electronegative atom and it is a nucleophile. This implies it loves the nucleus which is positively charged.

The Carbon atom attached to it is an electrophile because it has a lower electronegativity thereby leaving a partial positive charge on it. Therefore, the carbon will be attracted to the nucleophile.

4 0
2 years ago
Last science question!! Thank you all for the help!
AlladinOne [14]

Answer 11 moles of O2 and 10 moles of H2O

5 0
2 years ago
1. A 9.941 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 14.57 grams of CO2 an
salantis [7]

Answer:

1) empirical formula = CH2O ; the molecular formula = C2H402

2) empirical formule = C3H404 = Molecular formula

Explanation:

CxHyOz  → CO2 + H2O

⇒ So for CO2 we can say that for each 1 C-atom (12g) there are 2 O-atoms (32g).

This means the ratio is 12g C/ 32g O

CO2 has a molar mass of 44 g/mole

in 14.57 grams of CO2 there is: (12g Carbon / 44g CO2) * 14.57 g CO2 = 3.9736 g Carbon

⇒For H2O we can do the same:

The ratio of each element in H2O is: 2 H-atoms/ 1 O-atom  This is 2g H/ 16g O

In 5.966 g of water there are: [2 g H / 18 g H2O] * 5.966 g H2O = 0.6629 g H

⇒The original sample had 9.941 g of Sample - 3.9736 g of C - 0.6629 g of H =  5.3045 g of O

2) Calculate number of moles

C: 3.9736 g / 12.0 g/mol = 0.3311 mol

H: 0.6629 g / 1.0 g/mol = 0.6629 mol

O: 5.3045g / 16.0 g/mol =  0.3315 mol

3) Now we should divide each number of mole by the smallest number (0.3311) to find the proportion of the elements

C: 0.3311/ 0.3311= 1

H: 0.6629 / 0.3311 = 2

O: 0.3315/ 0.3311  = 1

This gives us the empirical formula of CH2O

4) Calculate the mass of the empirical formula

Molar mass of Carbon = 12g /mole

Molar mass of Hydrogen = 1g /mole

Molar mass of Oxygen = 16g /mole

mass of the empirical formule = 12 + 2*1 + 16 = 30g

5) Calculate the molecular formula

mass of molecular formule / mass of empirical formula = n

We have to multiply the empirical formula by n to get the molecular formula.

60 / 30 = 2 = n

If we multiply CH2O by 2 we'll get: C2H4O2

If we control this by calculating the molar mass:

2*12 + 4*1.01 + 2*16 = 60.04 g/mole

Then, the molecular formula is C2H4O2

CxHyOz  → CO2 + H2O

⇒ So for CO2 we can say that for each 1 C-atom (12g) there are 2 O-atoms (32g).

This means the ratio is 12g C/ 32g O

CO2 has a molar mass of 44 g/mole

in 17.97 grams of CO2 there is: (12g Carbon / 44g CO2) * 17.97 g CO2 = 4.90 g Carbon

⇒For H2O we can do the same:

The ratio of each element in H2O is: 2 H-atoms/ 1 O-atom  This is 2g H/ 16g O

In 4.905 g of water there are: [2 g H / 18 g H2O] * 4.905 g H2O = 0.545 g H

⇒The original sample had 14.16 g of Sample - 4.90 g of C - 0.545 g of H =  8.715g of O

2) Calculate number of moles

C: 4.90 g / 12.0 g/mol = 0.4083 mol

H: 0.545 g / 1.0 g/mol = 0.545 mol

O: 8.715g / 16.0 g/mol =  0.5447 mol

3) Now we should divide each number of mole by the smallest number (0.4083) to find the proportion of the elements

C: 0.4083/ 0.4083= 1

H: 0.545 / 0.4083 =1.33

O: 0.5447/ 0.4083  = 1.33

We should multiply everything by 3

This gives us the empirical formula of C3H4O4

4) Calculate the mass of the empirical formula

Molar mass of Carbon = 12g /mole

Molar mass of Hydrogen = 1g /mole

Molar mass of Oxygen = 16g /mole

mass of the empirical formule = 3*12 + 4*1 + 4*16 = 104

5) Calculate the molecular formula

mass of molecular formule / mass of empirical formula = n

We have to multiply the empirical formula by n to get the molecular formula.

104 / 104 = 1 = n

This means the empirical formula = molecular formula = C3H4O4

4 0
3 years ago
In negative adsorption the concentration of the solution is​
trapecia [35]

Answer:

This is called negative adsorption. Hence if the concentration of the adsorbate is less on the surface of the adsorbent than in the bulk, it is known as negative adsorption.

Explanation:

Hope this helps

7 0
2 years ago
How might you add warm water to the beaker of cold water?
Trava [24]
You get a tub of warm water and tip the warm water into the cold water
7 0
3 years ago
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